You are given an integer array sorted in ascending order (may contain duplicates), you need to split them into several subsequences, where each subsequences consist of at least 3 consecutive integers. Return whether you can make such a split.

Example 1:

Input: [1,2,3,3,4,5]
Output: True
Explanation:
You can split them into two consecutive subsequences : 
1, 2, 3
3, 4, 5

Example 2:

Input: [1,2,3,3,4,4,5,5]
Output: True
Explanation:
You can split them into two consecutive subsequences : 
1, 2, 3, 4, 5
3, 4, 5

Example 3:

Input: [1,2,3,4,4,5]
Output: False

Note:

The length of the input is in range of [1, 10000]

给一个升序排列的整数数(可能含有重复元素)，把这个数组拆分成几个至少含有3个整数的子序列，求是否可以拆分成功？（也就是能够全部拆分组成顺子）

解法：贪婪算法Greedy

Java:

public boolean isPossible(int[] nums) {
    Map<Integer, Integer> freq = new HashMap<>(), appendfreq = new HashMap<>();
    for (int i : nums) freq.put(i, freq.getOrDefault(i,0) + 1);
    for (int i : nums) {
        if (freq.get(i) == 0) continue;
        else if (appendfreq.getOrDefault(i,0) > 0) {
            appendfreq.put(i, appendfreq.get(i) - 1);
            appendfreq.put(i+1, appendfreq.getOrDefault(i+1,0) + 1);
        }   
        else if (freq.getOrDefault(i+1,0) > 0 && freq.getOrDefault(i+2,0) > 0) {
            freq.put(i+1, freq.get(i+1) - 1);
            freq.put(i+2, freq.get(i+2) - 1);
            appendfreq.put(i+3, appendfreq.getOrDefault(i+3,0) + 1);
        }
        else return false;
        freq.put(i, freq.get(i) - 1);
    }
    return true;
}

Python:

# Time:  O(n)
# Space: O(1)
class Solution(object):
    def isPossible(self, nums):
        """
        :type nums: List[int]
        :rtype: bool
        """
        pre, cur = float("-inf"), 0
        cnt1, cnt2, cnt3 = 0, 0, 0
        i = 0
        while i < len(nums):
            cnt = 0
            cur = nums[i]
            while i < len(nums) and cur == nums[i]:
                cnt += 1
                i += 1

            if cur != pre + 1:
                if cnt1 != 0 or cnt2 != 0:
                    return False
                cnt1, cnt2, cnt3 = cnt, 0, 0
            else:
                if cnt < cnt1 + cnt2:
                    return False
                cnt1, cnt2, cnt3 = max(0, cnt - (cnt1 + cnt2 + cnt3)), \
                                   cnt1, \
                                   cnt2 + min(cnt3, cnt - (cnt1 + cnt2))
            pre = cur
        return cnt1 == 0 and cnt2 == 0

Python:

def isPossible(self, nums):
        left = collections.Counter(nums)
        end = collections.Counter()
        for i in nums:
            if not left[i]: continue
            left[i] -= 1
            if end[i - 1] > 0:
                end[i - 1] -= 1
                end[i] += 1
            elif left[i + 1] and left[i + 2]:
                left[i + 1] -= 1
                left[i + 2] -= 1
                end[i + 2] += 1
            else:
                return False
        return True

C++:

class Solution {
public:
	bool isPossible(vector<int>& nums)
	{
		unordered_map<int, priority_queue<int, vector<int>, std::greater<int>>> backs;

		// Keep track of the number of sequences with size < 3
		int need_more = 0;

		for (int num : nums)
		{
			if (! backs[num - 1].empty())
			{	// There exists a sequence that ends in num-1
				// Append 'num' to this sequence
				// Remove the existing sequence
				// Add a new sequence ending in 'num' with size incremented by 1 
				int count = backs[num - 1].top();
				backs[num - 1].pop();
				backs[num].push(++count);

				if (count == 3)
					need_more--;
			}
			else
			{	// There is no sequence that ends in num-1
				// Create a new sequence with size 1 that ends with 'num'
				backs[num].push(1);
				need_more++;
			}
		}
		return need_more == 0;
	}
};

C++:　　

class Solution {
public:
    bool isPossible(vector<int>& nums) {
        unordered_map<int, int> freq, need;
        for (int num : nums) ++freq[num];
        for (int num : nums) {
            if (freq[num] == 0) continue;
            else if (need[num] > 0) {
                --need[num];
                ++need[num + 1];
            } else if (freq[num + 1] > 0 && freq[num + 2] > 0) {
                --freq[num + 1];
                --freq[num + 2];
                ++need[num + 3];
            } else return false;
            --freq[num];
        }
        return true;
    }
};